Preprint Article Version 4 Preserved in Portico This version is not peer-reviewed

P versus NP

Version 1 : Received: 1 August 2019 / Approved: 5 August 2019 / Online: 5 August 2019 (03:31:23 CEST)
Version 2 : Received: 26 November 2019 / Approved: 29 November 2019 / Online: 29 November 2019 (07:28:14 CET)
Version 3 : Received: 4 April 2020 / Approved: 6 April 2020 / Online: 6 April 2020 (12:57:55 CEST)
Version 4 : Received: 15 April 2020 / Approved: 16 April 2020 / Online: 16 April 2020 (10:17:03 CEST)
Version 5 : Received: 18 September 2020 / Approved: 19 September 2020 / Online: 19 September 2020 (09:51:02 CEST)

How to cite: Vega, F. P versus NP. Preprints 2019, 2019080037 (doi: 10.20944/preprints201908.0037.v4). Vega, F. P versus NP. Preprints 2019, 2019080037 (doi: 10.20944/preprints201908.0037.v4).

Abstract

$P$ versus $NP$ is considered as one of the most important open problems in computer science. This consists in knowing the answer of the following question: Is $P$ equal to $NP$? It was essentially mentioned in 1955 from a letter written by John Nash to the United States National Security Agency. However, a precise statement of the $P$ versus $NP$ problem was introduced independently by Stephen Cook and Leonid Levin. Since that date, all efforts to find a proof for this problem have failed. It is one of the seven Millennium Prize Problems selected by the Clay Mathematics Institute to carry a US 1,000,000 prize for the first correct solution. Another major complexity class is $\textit{P-Sel}$. $\textit{P-Sel}$ is the class of decision problems for which there is a polynomial time algorithm (called a selector) with the following property: Whenever it's given two instances, a "yes" and a "no" instance, the algorithm can always decide which is the "yes" instance. It is known that if $NP$ is contained in $\textit{P-Sel}$, then $P = NP$. In this paper we consider the problem of computing the sum of the weighted densities of states of a Boolean formula in $3CNF$. Given a Boolean formula $\phi$, the density of states $n(E)$ counts the number of truth assignments that leave exactly $E$ clauses unsatisfied in $\phi$. The weighted density of states $m(E)$ is equal to $E \times n(E)$. The sum of the weighted densities of states of a Boolean formula in $3CNF$ with $m$ clauses is equal to $\sum_{E = 0}^{m} m(E)$. We prove that we can calculate the sum of the weighted densities of states in polynomial time. The lowest value of $E$ with a non-zero density (i.e. $min_{E}\{E|n(E) > 0\}$) is the solution of the corresponding $\textit{MAX-SAT}$ problem. The minimum lowest value with a non-zero density from the two formulas $\phi_{1}$ and $\phi_{2}$ is equal to the minimum value between $E_{1}$ and $E_{2}$, where $E_{i}$ is the lowest value with a non-zero density of $\phi_{i}$ for $i \in \{1, 2\}$. Given two Boolean formulas $\phi_{1}$ and $\phi_{2}$ in $3CNF$ with $n$ variables and $m$ clauses, the combinatorial optimization problem $\textit{SELECTOR-3SAT}$ consists in selecting the formula which has the minimum lowest value with a non-zero density, where every clause from $\phi_{1}$ and $\phi_{2}$ can be unsatisfied for some truth assignment. We assume that the formula with the minimum lowest value with a non-zero density has the minimum sum of the weighted densities of states. In this way, we solve $\textit{SELECTOR-3SAT}$ with an exact polynomial time algorithm. Finally, we claim that this could be used for a possible selector of $3SAT$ and thus, $P = NP$.

Subject Areas

complexity classes; combinatorial optimization; polynomial time; reduction; logarithmic space; one-way

Comments (1)

Comment 1
Received: 16 April 2020
Commenter: Frank Vega
Commenter's Conflict of Interests: Author
Comment: The content was simplified. In this way, the abstract and content of the paper were changed and improved.
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